数列an,a1=1,a2=2,An+2=(An+An+1)/2,n为正整数
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数列an,a1=1,a2=2,An+2=(An+An+1)/2,n为正整数
数列an,a1=1,a2=2,An+2=(An+An+1)/2,n为正整数
数列an,a1=1,a2=2,An+2=(An+An+1)/2,n为正整数
a(n+2)=(a(n)+a(n+1))/2
两边减去a(n+1)
a(n+2)-a(n+1)=-1/2*(a(n+1)-a(n))
所以a(n+1)-a(n)是等比数列,a(2)-a(1)=1
a(n+1)-a(n)=(-1/2)^(n-1)
两边求和S(n+1)-a(1)-S(n)=(1-(-1/2)^n)/(3/2)
即a(n+1)=1+2/3-2/3*(-1/2)^n=5/3-2/3*(-1/2)^n
所以a(n)=5/3+4/3*(-1/2)^n
a(n+2)=[an+a(n+1)]/2
2a(n+2)=an+a(n+1)
2a(n+2)-2a(n+1)=an-a(n+1)
2[a(n+2)-a(n+1)]=an-a(n+1)
[a(n+2)-a(n+1)]/[a(n+1)-an]=-1/2
所以a(n+1)-an是以-1/2为比的等比数列
a(n+1)-an=(a2-a1)q^(n-1)
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a(n+2)=[an+a(n+1)]/2
2a(n+2)=an+a(n+1)
2a(n+2)-2a(n+1)=an-a(n+1)
2[a(n+2)-a(n+1)]=an-a(n+1)
[a(n+2)-a(n+1)]/[a(n+1)-an]=-1/2
所以a(n+1)-an是以-1/2为比的等比数列
a(n+1)-an=(a2-a1)q^(n-1)
a(n+1)-an=(2-1)(-1/2)^(n-1)
a(n+1)-an=(-1/2)^(n-1)
a(n+1)-an=(-1/2)^(n-1)
an-a(n-1)=(-1/2)^(n-2)
..........
a3-a2=(-1/2)^1
a2-a1=(-1/2)^0
以上等式相加得
a(n+1)-a1=(-1/2)^0+(-1/2)^1+...........+(-1/2)^(n-1)
a(n+1)-1=[1-(-1/2)^n]/[1-(-1/2)]
a(n+1)-1=2*[1-(-1/2)^n]/3
a(n+1)-1=2/3+(-1/2)^(n-1)/3
a(n+1)=5/3+(-1/2)^(n-1)/3
an=5/3+(-1/2)^(n-2)/3
收起
二人的方法没错。